Cost Minimization of a Multiple Section Power Cable Supplying Several Remote Telecom Equipment

Transcription

1 Cost Minimization of a Multiple Section Power Cable Supplying Several Remote Telecom Equipment Joaquim J Júdice Victor Anunciada Carlos P Baptista Abstract An optimization problem is described, that arises in telecommunications and is associated with multiple cross-sections of a single power cable used to supply remote telecom equipments The problem consists of minimizing the volume of copper material used in the cables and consequently the total cable cost Two main formulations for the problem are introduced and some properties of the functions and constraints involved are presented In particular it is shown that the optimization problems are convex and have a unique optimal solution A Projected Gradient algorithm is proposed for finding the global minimum of the optimization problem, taking advantage of the particular structure of the second formulation An analysis of the performance of the algorithm for given real-life problems is also presented Keywords: Nonlinear Optimization, Convex Programming, Energy Systems, Telecommunications 1 Introduction This work describes the formulation and resolution of an engineering optimization problem that arose in a project of the Instituto de Telecomunicações (Portuguese research institute for telecommunications) with the company Portugal Telecom [1] The problem consists of optimizing the multiple cross-sections of a single power cable used to supply remote telecom equipment Electric power transmission is usually operated at an almost constant voltage In fact, electric appliances and equipment require a constant supply voltage, named the nominal Departamento de Matemática, Universidade de Coimbra and Instituto de Telecomunicações, Portugal JoaquimJudice@coitpt Instituto Superior Técnico, UTL and Instituto de Telecomunicações, Portugal avaa@lxitpt V Anunciada passed away on September 29, 2007 Escola Superior de Tecnologia, Instituto Politécnico de Tomar, Portugal CarlosPerquilhas@aimesttiptpt 1

2 operating voltage In alternating current (AC) distribution networks, the nominal operating voltage is 230 V in Europe and 110 V in USA Those values are the root mean square value of a sinusoidal voltage that is a function of time and frequency (frequencies of 50 Hz in Europe and 60 Hz in USA) Electric appliances may accept only narrow variations of those voltages, typically tolerances of +10%, -15% If an electrical generator operates at a fixed voltage V and a distribution line with several nodes, where loads are connected, there are voltage drops along the line, described by Ohm s law (the voltage drop is proportional to the line impedance and line current, and line impedance is proportional to line length) However, when currents are at a maximum value in any one of the nodes, the voltage must be kept in the interval of acceptable voltages Consequently, power distribution is made with oversized copper or aluminum cables The maximum length of the distribution cables is limited, in order to avoid excessive voltage drops or excessive cross sections of the cables Usually the cable length is limited to be less than 2 km The cables cross sections are standardized considering the maximum current in the system and the maximum allowed voltage drops For larger distances, the distribution networks use transformers in order to increase operating voltage to 10 kv or 15 kv, reducing the line currents in the same proportion For very long distances or large power, the voltage increases up to 440 kv or more Power distribution operates at an almost constant voltage Consequently, power lines are interconnected with transformers in order to keep acceptable sizes of the cables Industry has standardized cables, maximum distances and operating voltages in order to obtain acceptable values for the system cost and voltage drops A different situation occurs when a small amount of energy is required very far from any possible supply point This happens on a motorway, far from any town where it is required to power a video camera and the radio transmitter The use of optic fibers in long distance telecommunications implies the use of electronic signal conditioning equipment placed along the fibers, in places that can be more than 100 km from the nearest power supply It is possible to construct a copper cable for energy supply along the fiber cable However, such long distances imply very large voltage drops or bulky and costly cables, assuming a constant voltage operation In the above mentioned project, the authors considered the possibility of having very large voltage drops in the cable, in order to avoid large cable costs This option is possible because telecom equipment require voltage regulators and storage batteries that can operate with voltage tolerances up to 50% When the power cable supplies several devices in different places along the cable, the cross section of each part of the cable should be optimized As shown in this paper, the model under consideration can be formulated as a nonlinear programming problem We are able to show that the constraint set of this program is bounded and the objective function is strictly convex on this constraint set Hence the optimization problem has a unique optimal solution, which can easily be found by a nonlinear program code, such as MINOS [14] By exploiting the structure of this optimization problem, it is also 2

3 possible to reformulate it as a strictly convex nonlinear program on a simplex This latter optimization problem can be efficiently processed by a Projected Gradient algorithm [3, 4, 5] that fully exploits the structure of the constraint set Computational experience with small instances of the problem illustrates the validity of the formulation for its purpose and of the Projected Gradient algorithm to process the associated nonlinear program The organization of the paper is as follows In section 2 the model and its formulation are introduced A heuristic procedure used in [1] is reviewed in section 3 to find a feasible solution to the associated optimization problem The solution of the nonlinear program under consideration is fully discussed in section 4 The alternative formulation with simplex constraints and the Projected Gradient algorithm for its solution are described in sections 5 and 6 Finally some computational experience and conclusions are reported 2 Model Description A long cable with two conducting wires is supplied at one end by a constant voltage generator with a known voltage v I The cable is constituted by n sections The cross-sections of the conductors may differ from section to section (although in each section of the cable the cross-section of both conducting wires should be the same) If l i is the length of each section i (i = 1, 2,, n), then the total length of the cable is n i=1 l i Besides the voltage generator and the n sections, we assume the existence of n nodes along the cable, each representing an electrical load Section i connects two consecutive nodes i 1 and i, thus Section i Node i 1 Node i where i = 1, 2,, n (and node 0 coincides with the voltage generator) Therefore node i connects two consecutive sections i and i + 1: Section i Section i + 1 Node i where i = 1, 2,, n 1, with the exception of node n The load in each node i is known and is characterized by its constant power load p i The load current l c i in each node i depends on the voltage v i in that node, according to The voltage decrease v i along each section i ( v i law [10], l c i = p i v i, i = 1, 2,, n (1) = v i 1 v i ) is, according to Ohm s v i = v i 1 v i = r i c i, (2) 3

4 where r i and c i are the resistance of the cable and the current in section i, i = 1, 2,, n As the cable is made of two conducting wires, the resistance of each cable section is twice the resistance of each conductor in that section Hence r i = 2 ρ l i s i, (3) where ρ is the material resistivity, l i is the length of the section, and s i is the area of the cross section The current c i in section i is the sum of the currents in the loads at nodes i, i + 1,, n c i = n lc j = n v j (4) Consequently the current in section i (i = 1, 2,, n) depends on the voltage in all later nodes of this section, as the power loads p i are constant Now, using the expressions of r i and s i in (3) and (4), we obtain from (2) Therefore for i = 1, 2,, n v i 1 v i = 2 ρ l i n s v j (5) i v i = v i 1 2 ρ l i s i n v j (6) Another data item is the voltage v n in the last node n of the cable Considering that the voltage v 0 of node 0 coincides with the initial voltage v I, then where a is a positive given constant satisfying 1 2 a < 1 1 v n = a v 0, (7) The volume of each section i of the cable is equal to 2 l i s i Therefore, the total volume of cable material is V = n i=1 2 l i s i (8) As the price of the cable is almost proportional to its volume, its total cost C is given by C = c V, where c is a given positive constant C = c n i=1 2 l i s i = n i=1 2 c l i s i (9) The objective of the problem is to determine the cross-section area s i of the conducting wires in each section that minimizes the total cost of the cable This is equivalent to minimizing its volume V The values p i and l i, i = 1, 2,, n, as well as v 0, v n, a, ρ and c are data of the optimization problem, and we have to find the cross-section areas s i, i = 1, 2,, n, that minimize C (or equivalently V ) The problem also contains the variables v 1, v 2,, v n 1, which are related to s i, i = 1, 2,, n by (6) 1 It is possible to show that if a < 1, the total volume of the cable increases as the voltage vn decreases [7] 2 4

5 The model described can then be formulated as the following nonlinear optimization problem: Minimize subject to n i=1 2 c l i s i (10) v i = v i 1 2 ρ l i n, i = 1, 2,, n, s i v j (11) v n = a v 0, (12) v i 1 > v i, i = 1, 2,, n, (13) s i, v i > 0, i = 1, 2,, n, (14) where the power loads p i (i = 1, 2,, n), the lengths l i (i = 1, 2,, n), the voltages v 0 and v n, and the constants a, ρ and c, are data, and the cross-sections s i (i = 1, 2,, n) and the voltages v 1, v 2, v n 1 are the variables of the optimization problem The objective in (10) is the minimization of the total cost of the cable or equivalently of its total volume 3 An Initial Feasible Solution for the Optimization Problem As stated in [1], a first feasible solution for the optimization problem can be constructed by assuming that in each cable section i, i = 1, 2,, n, the cross section s i is proportional to the current c i Thus s i = S c i, and from (4), s i = S n lc j, with S > 0 The process of finding such a solution reduces to the computation of the proportionality constant S Suppose that all the loads along the cable are concentrated in a unique node, so that the cable is made of a single section with length L = n i=1 l i and by two nodes, one at each end, namely the voltage generator and the current node This last node concentrates the electrical loads, whence it has a load power P = n i=1 p i Since the voltage in this node is a v 0, we can easily compute the load current c in the last node, the resistance r in the section, and the cross-section s in the conducting wires of that section by c = P a v 0, (15) Therefore the reference cross-section S is defined by r = v 0 a v 0 = a (1 a) v2 0, (16) c P s = 2 ρ L 2 ρ LP = r a (1 a) v0 2 (17) S = s c = The cross-sections s i are computed from 2 ρ L (1 a) v 0 (18) s i = S n lc j, i = 1, 2,, n (19) 5

6 The voltages v i can now be obtained recursively from (6) by v i 1 = v i + 2 ρ l i s i n v j, i = 1, 2,, n (20) In the particular case where p 1 = p 2 = = p n, with v n = v 0, the voltages can be 2 determined as follows: v i 1 v i = l i n i=1 l (v 0 v n ), i = 1, 2,, n (21) i Then by (20) the cross-sections satisfy n v j s i = 2 ρ l i, i = 1, 2,, n (22) v i 1 v i Computational experience (section 7) shows that this process finds a feasible solution that is in general a tight upper bound to the global minimum value of the optimization problem It is also important to note that, since s i = S n lc j, equations (5) can be rewritten as v i = v i 1 v i = 2 ρ S l i This means that for the feasible solution obtained through this heuristic method, the voltage decrease v i in each section i is proportional to the length l i of this section 4 Solution of the Nonlinear Optimization Problem Consider again the model described in the section 2 Looking carefully at this formulation, we note that the constraints s i, v i > 0, i = 1, 2,, n, are redundant In fact, according to (12) and (13), we have v 0 > v 1 > v 2 > > v n = a v 0 Since a and v 0 are positive, then v i a v 0 > 0, for all i = 1, 2,, n As v i 1 v i > 0, ρ > 0, l i > 0 and > 0, for all j, then s i > 0, for all i = 1, 2,, n Then the variables s i can be assumed to be unrestricted in sign and can be eliminated from the problem to obtain the following nonlinear program in the variables v i : NLP: Minimize v R n n 4 ρ c n v j i=1 l2 i (23) v i 1 v i subject to v i 1 > v i, i = 1, 2,, n, (24) v n = a v 0, (25) where v 0, a, ρ, c, l i, are positive data After a global minimum is obtained for this optimization problem NLP, then the cross-section areas s i can be found from (11) The feasible set of this program is not closed, whence the Weierstrass Theorem [2] cannot be used to guarantee the existence of a minimum To overcome this difficulty, we consider the additional variables z i, defined by z i = v i 1 v i, i = 1, 2,, n, and the nonlinear program: 6

7 n NLP1: Minimize f(z, v) = 4 ρ c n v j i=1 l2 i (26) z i subject to z i v i 1 + v i = 0, i = 1, 2,, n, (27) z i ε, i = 1, 2,, n, (28) v n = a v 0, (29) where ε is a positive and small real number (in practice ε = 10 6 ) As for each i = 1, 2,, n, v i 1 > v i z i = v i 1 v i > 0, then we guarantee that the function f of NLP1 has a global minimum on the set K = {(z, v) R n n satisfying (27) (29)} Consider now the objective function of NLP1, the n functions f i defined by and the sets f i (z i, v i,, v n ) = n v j z i, i = 1, 2,, n (30) K i = {(z i, v i,, v n ) R n i+2 : z i > 0; v j > 0, j = i,, n}, i = 1, 2,, n (31) Therefore for all (z, v) K, f(z, v) = 4 ρ c n i=1 l2 i f i(z i, v i,, v n ) (32) Then f is strictly convex on K, as the following property holds: Theorem The function f i ( i = 1, 2,, n ) defined by (30) is strictly convex on K i Proof: The gradient of f i at each point of K i always exists and is given by The Hessian of f i is 1 zi 2 f i (z i, v i,, v n ) = 7 ( n v j 1 z i ( pi v 2 i 1 z i ( pn v 2 n ) ) )

8 where and A 2 is given by A 2 = 2 f i (z i, v i,, v n ) = A 1 + A 2, ( 1 A 1 = diag zi 3 1 z 3 i ( n p i zi 2 v2 i p i+1 zi 2 v2 i+1 p n z 2 i v2 n ( n ) v j p i zi 2 v2 i p i z i vi 3 0 ), v j ) p i p n z i vi 3,, z i vn 3 p i+1 z 2 i v2 i+1 p n z 2 i v2 n 0 0 p i+1 z i v 3 i p n z i v 3 n Now A 2 can be written as follows: [ α A 2 = v n i+1 and where α > 0, v ( ) pi p i+1 p n D = diag z i vi 3, z i vi+1 3,, z i vn 3 Since all diagonal elements of D are positive and the Schur Complement (A 2 D) of D in A 2 [6] is zero, then A 2 is Symmetric Positive Semi-Definite As A 1 is Symmetric Positive Definite, then 2 f i (z i, v i,, v n ) is Symmetric Positive Definite in K i [6] and f i is strictly convex on K i v T D ], Since K is compact and f is strictly convex on K, then f has a unique stationary point on K, which is exactly its unique global minimum [2] As a stationary point of f on K may be found by a local nonlinear programming algorithm, it is easy to obtain the unique global minimum of the function on K and thus to solve the optimization problem 5 An Alternative Formulation on the Simplex Consider the program NLP1, introduced in the previous section We can write the constraints of the program as follows: Lv + z = b, v n = a v 0, z i ε, i = 1, 2,, n, 8

9 where L = R n n, b = v R n, z = z 1 z 2 z 3 z n R n As L is a triangular nonsingular matrix, we can eliminate the variables v i by solving Lv = b z Then Furthermore v i = v 0 i k=1 z k, i = 1, 2,, n (33) v n = a v 0 v 0 (z 1 + z z n ) = a v 0 z 1 + z z n = (1 a) v 0 n j=1 z j = (1 a) v 0 Then NLP is equivalent to NLP2: Minimize f(z) = 4 ρ c n li 2 i=1 z i subject to e T z = (1 a) v 0, z i ε, i = 1, 2,, n, ( n v 0 j k=1 z k ) where e n is a vector of ones The feasible set of this program is given by K = {z R n : e T z = (1 a) v 0 ; z i ε, i = 1, 2,, n} (34) Hence K is a compact set and by Weierstrass Theorem, the function f has a global minimum on K Moreover, the function of NLP2 is obtained from the function of NLP by a linear transformation with a nonsingular matrix Therefore this function f is strictly convex on K [13] Hence the computation of the unique global minimum of NLP2 may be done by a local nonlinear programming method In the next section we introduce a Projected Gradient algorithm that finds the unique stationary point of f on K by taking advantage of the structure of the program NLP2 The original variables v i and s i can be obtained afterwards by using formulas (33) and (11) 9

10 6 A Projected Gradient Algorithm for the Nonlinear Program on the Simplex Consider the program NLP2 introduced in the previous section and its constraint set K given by (34) The projected gradient of f in the point z K is defined as the vector in g(z) = P K (z η f(z)) z, (35) n where P K ( ) is the orthogonal projection in K and η > 0 [3] It is known that for a given η > 0, g(z) = 0 if and only if z is a stationary point of f on K [3] The Projected Gradient algorithm is a modification of the Steepest Descent algorithm [8] in which the iterates are forced to stay in K by a projection [4] It can be shown [3, 5] that this algorithm possesses global convergence towards a stationary point under reasonable hypotheses The steps of the algorithm are presented below Projected Gradient Algorithm for NLP2: Step 0: Find an initial solution z K Step 1: (Computation of Search Direction) Compute the direction d n by d = y z, where y = P K (z η f(z)), with η > 0 Step 2: (Stopping Test) If d 2 < tol, then stop: z is a stationary point of f on K Otherwise, go to Step 3 Step 3: (Armijo Criterion) Find α + such that f(z + α d) f(z) + β α f(z) T d, with 0 < β < 1 Step 4: (Iterate Updating) Update the current solution z z + α d and return to Step 2 Next, we discuss some details of the algorithm 10

11 (I) Computation of the Gradient of f By simple manipulation, it is possible to write f(z) as follows: f(z) = = 4 ρ c k lj 2 p n k j=1 z j k=1 k v 0 j=1 z j Therefore it is easy to obtain the following formulas for the components of the gradient vector f(z): i = 1,2,, n f(z) z i = 4 ρ c n k=i k lj 2 p k j=1 z j (v 0 k j=1 z j) 2 l2 i z 2 i n k=i p k v 0 k j=1 z j (36) (II) Computation of the Projection y = P K (z η f(z)) 1 Find u = z η f(z) using (36) to compute f(z) 2 The vector y is the unique optimal solution of the strictly convex quadratic program 1 Minimize y R n 2 u y 2 2 subject to e T y = (1 a) v 0, y i ε, i = 1, 2,, n But u y 2 2 = (u y) T (u y) = u T u 2 u T y + y T y and therefore this program is equivalent to Minimize z R n where b 0 = (1 a) v 0 and q = u q T y yt y subject to e T y = b 0, y i ε, i = 1, 2,, n, There are several algorithms described in the literature to process this kind of quadratic programs [9, 15, 12] Among these processes, the Block Pivotal Principal Pivoting (BLOCK) Algorithm described in [12] is strongly polynomial and very efficient in practice The steps of this process are presented below 11

12 BLOCK Algorithm: Step 0: Let F = { 1, 2,, n } Step 1: Compute λ = b 0 + i F q i F Step 2: (Stopping test) Let H = {i F : q i + λ > ε} If H = φ, stop: the vector ε, if i F z = (z i ) i = 1,2,, n, with z i = (q i + λ), if i F is the unique optimal solution of the quadratic program Otherwise set F = F H and go to Step 1 (III) Computation of the Optimal Values for the Variables v i and s i Let z = (z i ) i = 1,2,, n R n be the optimal solution of the problem NLP2 The variables v i, i = 1, 2,, n, of the original problem satisfy Lv = b z Therefore v i = v i 1 z i, i = 1, 2,, n ( v 0 = v 0 ) (37) Furthermore, (11) provides the values of the variables s i : s i = 2 ρ l i n 7 Computational Experience z i v j, i = 1, 2,, n The computational experience presented in this section was performed using a PC with 3 Ghz Pentium 4 processor and 2 Gb RAM memory, running Linux 2610 The active set code MINOS [14] (one of the nonlinear solvers of GAMS) has been used for processing both the nonlinear programs NLP1 and NLP2 Furthermore, the Projected Gradient algorithm for the program NLP2 was implemented in Fortran 77 [11], using the Gnu Fortran (g77) compiler, version 343, with the options -02 -malign-double -funroll-loops Running times presented in this section are given in CPU seconds, excluding inputs and outputs The times for the Projected Gradient algorithm were measured by the etime() function The results reported in this section are concerned with two actual instances of the problem found in practice, the first (Example 1) presenting reduced voltage decrease and the second 12

13 (Example 2) presenting high voltage decrease The data for these problems are stated below and in Tables 1 and 2: n = 40; ρ = 20 (Ω Km 1 mm 2 ); c > 0 Example 1 (reduced voltage decrease) Example 2 (high voltage decrease) V 0 = 500 V V 0 = 260 V a = 06 V 40 = 300 V V 40 = 190 V a 0731 i length l i (Km) power p i (W) length l i (Km) power p i (W) Table 1: Problem data for Examples 1 and 2 13

14 Example 1 (reduced voltage decrease) Example 2 (high voltage decrease) i length l i (km) power p i (w) length l i (km) power p i (w) Table 2: Problem data for Examples 1 and 2 (continuation) For the first example the cross-section value is S = 9680 (with L = 4840 and P = 7750), and for the second example, is S = 8811 (with L = 1542 and P = ) The heuristic procedure described in section 3 has found the following solutions: For Example 1: v = (v i ) i = 0, 1,, 40 = (500000, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ); s = (s i ) i = 1, 2,, 40 = (195930, , , , , , , , , , , , , , , , , , , , , , , 97987, 96685, 95382, 94077, 92768, 91458, 88823, 83493, 78095, 64396, 50518, 35287, 32104, 25730, 19347, 17744, 16133) 14

15 with total cost C = c For Example 2: v = (v i ) i = 0, 1,, 40 = (260000, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ); s = (s i ) i = 1, 2,, 40 = ( , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 53332) with total cost C = c As mentioned earlier, each of these values is an upper bound for the optimal value given by the nonlinear programs NLP1 and NLP2 The unique optimal solution of those programs found by MINOS is given below: For Example 1: v = (vi ) i = 0, 1,, 40 = (500000, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ); s = (s i ) i = 1, 2,, 40 = (156856, , , , , , , , , , , , , , , , , , , , , , , , , , , , 99429, 97521, 93586, 89506, 78755, 67253, 53127, 49982, 43473, 36549, 34720, 32831) 15

16 For Example 2: v = (vi ) i = 0, 1,, 40 = (260000, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ); s = (s i ) i = 1, 2,, 40 = ( , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ) For both examples, we have used as an initial point for MINOS the default choice given by the code and the solution computed by the heuristic procedure The values of the variables v i show that the constraints z i ε ( ε = 10 6) are inactive at the optimal solutions of both problems Tables 3 and 4 report the numerical results under these two choices and lead to the conclusion that it is recommended to start with the feasible solution given by the heuristic procedure It is also important to add that Formulation NLP1 is preferable if an active-set method such as MINOS is employed to process the nonlinear program The reason for this arises from the more complex formulas of the objective function in the Formulation NLP2 Example 1 Initial Solution Formulations By default (MINOS) Heuristic Solution Iterations CPU Cost Iterations CPU Cost NLP c c NLP c c Table 3: Comparative performance of the formulations NLP1 and NLP2 for Example 1, solved by GAMS/MINOS The implementation of the Projected Gradient algorithm for problem NLP2 requires a choice for the parameters β, θ and η used by the procedure For both examples, some tests 16

17 Example 2 Initial Solution Formulations By default (MINOS) Heuristic Solution Iterations CPU Cost Iterations CPU Cost NLP c c NLP c c Table 4: Comparative performance of the formulations NLP1 and NLP2 for Example 2, solved by GAMS/MINOS have been performed with the algorithm in order to make a sensitivity analysis for these parameters The best choices for the Projected Gradient algorithm have been achieved with β = 10 4, θ = 1 2, η = 10 2 for Example 1 and β = 10 4 (or β = 10 2 ), θ = 1 10, η = 10 3 for Example 2 Table 5 indicates the performance of the Projected Gradient algorithm (with tol = 10 6 ) for the best choice of the parameters and for two initial solutions These computational results show that the Projected Gradient algorithm is efficient to process the nonlinear program NLP2 The number of iterations required by the projected gradient method is slightly bigger than those of the active-set method (MINOS) displayed in tables 3 and 4 However, the algorithm performs faster than MINOS Another interesting conclusion is that the use of the initial solution given by the heuristic procedure does not seem to help for this method As the Projected Gradient algorithm fully exploits the structure of the nonlinear program and requires minimal storage, we believe that this method can be useful for the solution of larger instances of this optimization model Example 1 Initial Solution Iterations CPU Cost z i = b 0 40, i = 1, 2,, c Heuristic Solution c Example 2 Initial Solution Iterations CPU Cost z i = b 0 40, i = 1, 2,, c Heuristic Solution c Table 5: Best performances of the Projected Gradient algorithm The unique optimal solution found by the algorithms gives about 3% savings for Example 1 and about 12% savings for Example 2, when compared with the heuristic solutions used in 17

18 practice in the project reported in this paper We believe that more substantial savings can be achieved for larger instances of the energy model if the unique optimal solution of NLP1 or NLP2 is used instead of the solution that is nowadays employed in this project 8 Conclusion This paper describes formulations and solutions of an engineering optimization problem in telecommunications area, that consists of optimizing the cross section of conducting energy cables, used to supply remote telecom equipments, in order to minimize the volume of copper material used in the cables and consequently the cost Two alternative formulations for the problem have been introduced and a Projected Gradient algorithm was proposed for the second formulation, taking advantage of its particular structure Computational experience with an implementation of this algorithm has shown that the proposed methodology is efficient in practice We believe that this type of model and the Projected Gradient algorithm can also be useful in other engineering models used in energy supply, piping and fluid distribution with a similar structure, particularly when the number of nodes is quite large References [1] Victor Anunciada Cost minimization of a multiple section energy cable supplying remote telecom equipments Technical report, Instituto de Telecomunicações, Portugal, 2003 [2] M Bazaraa, H Sherali, and C Shetty Nonlinear Programming: Theory and Algorithms John Wiley and Sons, New York, 1993 [3] D Bertsekas Nonlinear Programming Athena Scientific, Massachusetts, 1999 [4] D P Bertsekas On the Goldstein-Levitin-Polyak gradient projection method IEEE Transactions on Automatic Control, 21: , 1976 [5] E G Birgin, J M Martinez, and M Raydan Nonmonotone spectral projected gradient methods on convex sets SIAM Journal on Optimization, 10: , 2000 [6] R W Cottle, J S Pang, and R E Stone The Linear Complementarity Problem Academic Press, New York, 1992 [7] Raymond DeCarlo Linear Circuit Analysis second ed, Pen-Lin Min, Oxford, 2001 [8] J E Dennis and R B Schnabel Numerical Methods for Unconstrained Optimization and Nonlinear Equations SIAM, Philadelphia, 1996 [9] J P Dussault, J A Ferland, and B Lemaire Convex quadratic programming with one constraint and bounded variables Mathematical Programming, 36:90 104,

19 [10] R P Feynman, R B Leighton, and M Sands The Feynman Lectures On Physics Addison-Wesley Publishing Company, New York, 1963 [11] Maximilian E Hehl Linguagem de Programação Estruturada FORTRAN 77 McGraw- Hill, São Paulo, 1987 [12] J J Júdice and F M Pires Solution of large-scale separable strictly convex quadratic programs on the simplex Linear Algebra and its Applications, 170: , 1992 [13] B Martos Nonlinear Programming: Theory and Methods North Holland, New York, 1975 [14] B A Murtagh and M A Saunders MINOS 51 user guide Technical report, Department of Operations Research, Stanford University, 1987 [15] P M Pardalos and N Kovoor An algorithm for a singly constrained class of quadratic programs subject to upper and lower bounds Mathematical Programming, 46: ,